Question Number 16691 by Tinkutara last updated on 25/Jun/17
![The horizontal range of a projectile is R and the maximum height attained by it is H. A strong wind now begins to blow in the direction of horizontal motion of projectile, giving it a constant horizontal acceleration equal to g. Under the same conditions of projection, the new range will be (g = acceleration due to gravity) [Answer: R + 4H]](https://www.tinkutara.com/question/Q16691.png)
$$\mathrm{The}\:\mathrm{horizontal}\:\mathrm{range}\:\mathrm{of}\:\mathrm{a}\:\mathrm{projectile} \\ $$$$\mathrm{is}\:{R}\:\mathrm{and}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{height}\:\mathrm{attained} \\ $$$$\mathrm{by}\:\mathrm{it}\:\mathrm{is}\:{H}.\:\mathrm{A}\:\mathrm{strong}\:\mathrm{wind}\:\mathrm{now}\:\mathrm{begins}\:\mathrm{to} \\ $$$$\mathrm{blow}\:\mathrm{in}\:\mathrm{the}\:\mathrm{direction}\:\mathrm{of}\:\mathrm{horizontal} \\ $$$$\mathrm{motion}\:\mathrm{of}\:\mathrm{projectile},\:\mathrm{giving}\:\mathrm{it}\:\mathrm{a}\:\mathrm{constant} \\ $$$$\mathrm{horizontal}\:\mathrm{acceleration}\:\mathrm{equal}\:\mathrm{to}\:{g}. \\ $$$$\mathrm{Under}\:\mathrm{the}\:\mathrm{same}\:\mathrm{conditions}\:\mathrm{of}\:\mathrm{projection}, \\ $$$$\mathrm{the}\:\mathrm{new}\:\mathrm{range}\:\mathrm{will}\:\mathrm{be} \\ $$$$\left({g}\:=\:\mathrm{acceleration}\:\mathrm{due}\:\mathrm{to}\:\mathrm{gravity}\right) \\ $$$$\left[\boldsymbol{\mathrm{Answer}}:\:{R}\:+\:\mathrm{4}{H}\right] \\ $$
Answered by ajfour last updated on 25/Jun/17
![time of flight doesn′t change, T=((2u_y )/g) = ((2usin θ)/g) new Range is R′= (ucos θ)T+(1/2)gT^2 = R+4[(1/2)g((T/2))^2 ] = R+4H .](https://www.tinkutara.com/question/Q16694.png)
$$\:\mathrm{time}\:\mathrm{of}\:\mathrm{flight}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{change}, \\ $$$$\:\mathrm{T}=\frac{\mathrm{2u}_{\mathrm{y}} }{\mathrm{g}}\:=\:\frac{\mathrm{2usin}\:\theta}{\mathrm{g}} \\ $$$$\mathrm{new}\:\mathrm{Range}\:\mathrm{is}\: \\ $$$$\:\:\mathrm{R}'=\:\left(\mathrm{ucos}\:\theta\right)\mathrm{T}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{gT}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\:\mathrm{R}+\mathrm{4}\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{g}\left(\frac{\mathrm{T}}{\mathrm{2}}\right)^{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:=\:\mathrm{R}+\mathrm{4H}\:. \\ $$
Commented by Tinkutara last updated on 25/Jun/17
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$